It is one of the most common algebra mistakes in any JEE paper, and it looks perfectly innocent on the page:

\sqrt{a^2 + b^2} \stackrel{?}{=} a + b

The idea behind it, usually half-conscious, is that the square root "undoes" the squares and leaves you with just a + b. After all, \sqrt{a^2} = a (for a \geq 0) and \sqrt{b^2} = b. Doesn't the radical just walk through the plus sign?

No. The square root does not distribute over addition. Never. And the easiest way to see this is with numbers.

A thirty-second check that breaks the myth

Pick a = 3 and b = 4.

\sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5.

But a + b = 3 + 4 = 7.

5 \neq 7. The two expressions are not equal — they differ by 2. Why: this single numerical counterexample is enough to disprove the general claim \sqrt{a^2+b^2} = a+b. If an identity is claimed for every a and b, exhibiting one pair where it fails is a complete refutation.

The trap is built in a very specific way. The student writes \sqrt{a^2 + b^2} and thinks: "Let me simplify." They then apply the (correct) fact that \sqrt{a^2} = a inside the expression, as if they could do it term by term. But the a^2 and b^2 are not separate radicals. They sit together under one radical sign, and there is no law that says you can pull the root across a sum.

The law that really applies

From Roots and Radicals, the two distribution laws for the square root are:

\sqrt{xy} = \sqrt{x} \cdot \sqrt{y} \quad (x, y \geq 0)
\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}} \quad (x \geq 0, y > 0)

Both of these are about multiplication and division, which are the operations inherited from exponents. There is no analogous rule for addition or subtraction:

\sqrt{x + y} \neq \sqrt{x} + \sqrt{y} \quad \text{in general.}

Why: the product rule comes from the power-of-a-product law (xy)^{1/2} = x^{1/2} y^{1/2}, which is a genuine exponent law. There is no corresponding exponent law for (x+y)^{1/2} — the binomial expansion of (x+y)^{1/2} is an infinite series, not a clean sum of powers. No distributivity, no short rule.

Why the mistake is seductive: the Pythagoras picture

There is also a geometric reason the mistake feels natural. In a right triangle with legs a and b, the hypotenuse has length \sqrt{a^2 + b^2}. The student's false identity \sqrt{a^2+b^2} = a + b is equivalent to claiming hypotenuse = sum of the two legs, which would mean the triangle is degenerate (flat).

A real right triangle has a hypotenuse strictly shorter than the sum of the other two sides — that is the triangle inequality. The correct relationship is

\sqrt{a^2 + b^2} \,<\, a + b \qquad \text{for } a, b > 0,

with equality only in the degenerate case when one of a, b is zero.

Right triangle showing that the hypotenuse is shorter than the sum of the legs A right triangle is drawn with horizontal leg of length three labelled a, vertical leg of length four labelled b, and hypotenuse of length five labelled square root of a squared plus b squared. A dashed line shows the path a plus b equal to seven that you would travel along the two legs to go from one corner to the other. A comparison panel on the right lists hypotenuse equals five, a plus b equals seven, and seven is greater than five. a = 3 b = 4 hypotenuse = 5 a = 3, b = 4 √(a² + b²) = 5 a + b = 7 5 < 7 ✓ The straight path (hypotenuse) is shorter than the two-leg detour. Always.
A $3$-$4$-$5$ right triangle. The hypotenuse $\sqrt{a^2+b^2} = 5$ is the straight-line path from one corner to the opposite; the leg-path of length $a + b = 7$ is the detour. $5 < 7$, and more generally $\sqrt{a^2+b^2} < a + b$ for any right triangle with both legs positive. The false identity claims the two paths have the same length.

The correct manipulation when you see \sqrt{a^2 + b^2}

The short answer: usually, you do not "simplify" it at all. \sqrt{a^2+b^2} is already as simplified as a generic expression of this shape can get.

In specific numerical cases, you evaluate inside the radical first:

\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13.

In symbolic expressions, you might be able to factor the sum inside the radical if it has a common factor — e.g., \sqrt{4a^2 + 4b^2} = \sqrt{4(a^2+b^2)} = 2\sqrt{a^2 + b^2} — but you cannot split the addition itself across the radical.

A diagnostic in one line

You are solving a geometry problem and at some step you write \sqrt{x^2 + 9}. Can you "simplify" this to x + 3?

Check with x = 4: \sqrt{16 + 9} = \sqrt{25} = 5, but x + 3 = 7. Not equal. So the simplification is wrong; leave the expression as \sqrt{x^2 + 9}.

Why: the same trap applies here. Underneath the radical is a sum, not a product. The square root does not distribute across the plus sign.

A related trap to watch for

(a + b)^2 is also not a^2 + b^2 — it is a^2 + 2ab + b^2, with a cross term. See Why (a+b)^2 Is Not a^2 + b^2 for the tile-area picture that explains where that cross term comes from and why it must be there.

Both mistakes have the same underlying cause: treating operations that do not distribute over addition as if they did. Squaring, square-rooting, and most non-linear operations behave cleanly with multiplication but never with addition. The safe instinct: before you move an operation across a plus sign, ask "does multiplication get similar treatment?" The square root example breaks that test the first time you try it.

Related: Roots and Radicals · Why (a+b)^2 Is Not a^2 + b^2 · Add Exponents When Multiplying, Multiply Them for a Power of a Power · Exponents Don't Distribute Over Addition · Is √16 Plus-or-Minus Four, or Just Four?